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L14
Integration
3
The
fundamental
Theorem
of
Calculus
Theorem
Suppose
that
the
function
f
is
continuous
on
an
interval
I
containing
the
point
a.
PART
2
.
Define
the
function
for
variable
x
=
I
by
11
Fix)
t
Safitsdt
,
x
=
I
.
Then
F
is
differentiable
on
I
,
and
Fix
=
fix)
.
So
F
is
an
antiderivative
of
f on
1
:
12)
*
)
,
"
fitsdt
=
fix)
.
Part
I
.
If
GN)
is
any
antiderivative
of
fixe
on
I
,
so
that
Gixs
=
fix)
on
1
,
then
for
any
b on
I
,
we
have
(3)
Saf(xdx
=
G1b)
-G(a)
.
Remark
:
0
Fix)
is
a
function
on
X
.
②
The
theorem
is
true
because
the
Mean
value
Theorem
of
integration
.
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③
The
theorem
is
the
most
impor tant
tool
for
the
calculation
of
integration
.
④
We
define
the
evaluation
symbol
Fixs/
!
f
=(b)
-
F1a)
,
and
use
the
symbol
/fixidx
for
the
indefinite
integral
,
i
.
e
.
general
antiderivative
of f
.
Then
B3)
becomes
S
!
fixsdx
=
((fix
dx))a
Proof
.
We
use
the
definition
of
derivative
F(xth)
-
FIX)
F(x)
=
h
=
(S
**
fitsat-
J
:
"
fits
at)
=
.
(
***
fits
at
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-
lim
↓
.
(xth-x)
ficl
with
c
+x
.
xth]
·
A
nto
Here
we
use
the
Mean
val ue
Theorem
for
integration
.
As
h
to
,
we
have
fixl
->
fix
.
so
above
limit
im
.
hf(x)
=
lim
f(x)
=
fix
,
c
->
X
4
where
we
use
the
fact
that
fixe
is
continuous
for
the
last
equality
.
So
we
checked
that
F(x)
=
f(x)
,
which
is
(2)
.
To
check
PART
I
,
we
obser ve
that
G(x)
=
f(x)
,
so
F(x)
=
DIX)
+
C
for
some
constant
C
.
So
fitdt=
FIx)
=
GIX)
+C
.
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Put
x
=
a
into
above
,
0
=
(i
fit)d
+
=
F(a)
=
G(a)
+
C
=>
c
=
-
G1a)
.
so
S
.
fitdt
=
G(x)
+)
=
G(x)
-
G(a)
.
Let
x=b
,
we
now
proved
(3)
(a
fit) dt
=
a(b)
-
G(a)
=
G(x)/a
-
i
-
Example
.
Find
the
area
for
the
region
=
41x
.
Y
GIRY)8
*
Y
*
Sinx
3
.
solution
.
Area
=I
*
sinxdx=7c0sx>
·
xY
=
f(0)x)
-
-10sO)
A
=
Sinx
↑
-
=
-
(
-
1)
+
1
=
2
.
B
<X
X
I
FcosX)'=
sinx
1
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Example
.
Find
the
derivatives
of
the
following
functions
:
cal
Fix
=
/
2
-
*
dt
,
(b)
G(x)
=
x
5x
e-tdt
,
·
14
14
Hix
=
I
e-tat
.
solution
.
We
ur e
the
fundamental
theorem
.
(a)
F(x)
=
- S
,
e
-tdt
So
Fix
=
-
(S
,
e
+
dt)
=
-
e
-
x
-
-
T
fundamental
theorem
for
fix) = e
*
(a)
<Method
2)
FIx)
=
-S
,
e
+it =/
,
)
-
et]
dt
we
use
fundamental
theorem
for
fix=-e
*
then
Fix)
=
-
e
-
*
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(b)
The
calculation
is
given
by
production
rule
for
derivative
.
(ax))
=
x2
)
.
et't
+
x2)
tat)'
·
5
=
2x
.
j
et'at
+
x2
e-
15x
-
chain
rule
x
e-tdt
-
1
*
et'dt
(2)
HIx
=
So
(Hixs)'
=
exe
*
==
x
e
·
2X
-
-
x
*
=
3x
x
-
2x
-
e
*
(
*
*
fitsdt
=
fit)
+
=
qx
.
9ix)
=
f(9
,
x))
.
gix)
.
*
-
fitd+=
f(gx)gix
-
f(hix)hix)
.
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Inte gration
Table
Polynomail
type
:
(xadx
=
+x
**
+
c(a+
-
1)
/x
+
dx
=
Jedx
=
en(x)
+
c
Trigeometric
type
:
/
sin(ax)
dx
=
-
I cos(ax)
+
> (a
= 0)
S
coslaxsdx
=
sin(axs
+
> (aF0)
I
secsaxidx
=
↓
tan laxs
+
>
(ato)
Hyper
trigeometuc
type
:
/sinh(axsdx
=
cosh(axs
+
<
(a
F0)
Scosh(ax)
&x
=
sinh
(ax)
+
>
(a
Fo)
Inverse
of
trigeometric
type
:
S
Idx
=
sin"
( )
+
>
(a
<o)
a
-
x2
I
artar
dx
=
I
tan")
+
C
.
Exponential
type
sea"dx=
e4+C
.
(b
**
dx
=
a
!s
b
**
+
C
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chain
rule
and
method
of
substitution
Recall
the
chain
rule
Pog)(x)
=
&(f(9
/
x)))
=
fig
,
x)
-
gix)
.
So
we
integrate
above
equation
and
get
(f(gix)
·
gixdx
=
fig,x)
+
C
We
let
u=
gixs
.
Then
du
d
X
=
giX)
.
so
in
differential
form
du=
gix
dx
.
If'(9(x))
-
gixdx
=
/fi(u)du
=
f(u)
+
C
=
f(9
1
x))
+
C
.
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Example
I
sin
(3
enx)
dx
X
solution
.
Let
U
=
ex
.
Then
du=
-
dx
/Sin
(3enx)
dx
=
/
Sin
(3
us
du
=S
sin
(3
us
d (34)
=
(d)-cossn)
=-
ws(ul
+
C
=
cos(1x)
+
C
i
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⑤Integration
by
parts
suppose
that
U(X)
and
VIx>
ar e
two
differentiable
functions
.
The
product
Rule
=>
*
(UIx)
.
X(x))
=
Mix a
+
xix)
Y
By
integrating
both
side
and
moving
terms
,
we
obtain
Jucx>
x
=
ux)
.
vix)
-
/vix)
d
*
We
also
write
above
by
SUNV=U
.
V-Svdu
.
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Example
I
x
edx
solution
Let
U
=
X
dy
=
exdx
=
de x
(xexdx
=
/
x d(eY)
=
Judy
=UV-SvdU
=
x
.
ex-
Sedx
=
x
.
e
-
e
+
C
I
Example
.
Seuxdx
=
JUdy
U=eux
.
V
=
x
=
U
.
X-SVdU
=
(nx)
.
X-
X
-
I
dx
=x1x
-
x
+
C(x
>
0)
1
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Summary
of
Today
:
*
)
fitdt
f(x)
mean
val ue
Theorem
Jaf(t)at=
(d)
/
computation
Tool
in
definite
integral
or
antiderivative
chain
rule
("composition
rule")
(f(gix)
·
gixdx
=
fig,x)
+
C
Integration
by
part
("production
rule"
(
Sudy
=
n
.
V-Sudu
Integration
table
.
